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eimsori [14]
3 years ago
11

How does the length and width of a flute affect its sound?

Physics
1 answer:
Butoxors [25]3 years ago
4 0

https://www.yamaha.com/en/musical_instrument_guide/flute/mechanism/mechanism002.html

Your welcome!

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A wave With wavelength 20 m has a frequency of 12 Hz what is the waves speed
pentagon [3]

Answer:

240m/s

Explanation:

The equation to calculate is wavelength= velocity/ frequency so to find the velocity you would have to multiply frequency by wavelength.

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Need help y’all ASAP please...physics
dolphi86 [110]

Answer:

t = 3/8 seconds

Explanation:

h=-16t^2 - 10t+6

h= 0 when it hits the ground

0=-16t^2 - 10t+6

factor out a -2

0= -2(8t^2 +5t -3)

divide by -2

0 = (8t^2 +5t -3)

factor

0=(8t-3) (t+1)

using the zero product property

8t-3 = 0    t+1 =0

8t = 3         t= -1

t = 3/8     t= -1

t cannot be negative  ( no negative time)

t = 3/8 seconds

3 0
2 years ago
The history of Sir Isaac Newton’s study of gravity does which of the following?
ipn [44]

A. Illustrates how a theory becomes a law

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3 years ago
Read 2 more answers
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vodka [1.7K]

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3 electron hai bro of puch mujhe sab aata h

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Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter
stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a

where,

m_c = mass of copper = 227 g

m_w = mass of water = 844 g

m_a = mass of aluminum = 155 g

C_c = specific heat capacity of calorimeter = 385 J/kg.°C

C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

8 0
2 years ago
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